MHB Berk's question via email about an antiderivative

Prove It
Gold Member
MHB
Messages
1,434
Reaction score
20
Evaluate $\displaystyle \begin{align*} \int{ \mathrm{e}^{-2\,x}\cos{(3\,x)}\,\mathrm{d}x} \end{align*}$

This requires using Integration By Parts twice...

$\displaystyle \begin{align*} I &= \int{\mathrm{e}^{-2\,x}\cos{ \left( 3\,x \right) } \,\mathrm{d}x} \\ I &= \frac{1}{3}\,\mathrm{e}^{-2\,x} \sin{(3\,x)} - \int{ -\frac{2}{3}\,\mathrm{e}^{-2\,x} \sin{(3\,x)}\,\mathrm{d}x } \\ I &= \frac{1}{3}\,\mathrm{e}^{-2\,x} \sin{(3\,x)} + \frac{2}{3} \int{ \mathrm{e}^{-2\,x} \sin{(3\,x)} \,\mathrm{d}x } \\ I &= \frac{1}{3}\,\mathrm{e}^{-2\,x} \sin{(3\,x)} + \frac{2}{3} \left[ -\frac{1}{3}\,\mathrm{e}^{-2\,x} \cos{(3\,x)} - \int{ \frac{2}{3}\,\mathrm{e}^{-2\,x} \cos{ \left( 3\,x \right) } \,\mathrm{d}x } \right] \\ I &= \frac{1}{3}\,\mathrm{e}^{-2\,x}\sin{ \left( 3\,x \right) } - \frac{2}{9}\,\mathrm{e}^{-2\,x} \cos{(3\,x)} - \frac{4}{9} \int{ \mathrm{e}^{-2\,x} \cos{(3\,x)} \,\mathrm{d}x } \\ I &= \frac{1}{3}\,\mathrm{e}^{-2\,x} \sin{ \left( 3\,x \right) } - \frac{2}{9} \,\mathrm{e}^{-2\,x} \cos{ \left( 3\,x \right) } - \frac{4}{9}\,I \\ \frac{13}{9}\,I &= \frac{3}{9}\,\mathrm{e}^{-2\,x} \sin{(3\,x)} - \frac{2}{9}\,\mathrm{e}^{-2\,x} \cos{ \left( 3\, x\right) } \\ I &= \frac{3}{13}\,\mathrm{e}^{-2\,x} \sin{(3\,x)} - \frac{2}{13}\,\mathrm{e}^{-2\,x} \cos{ \left( 3\,x \right) } \end{align*}$

Thus $\displaystyle \begin{align*} \int{ \mathrm{e}^{-2\,x} \cos{ \left( 3\,x \right) } \,\mathrm{d}x} = \frac{3}{13}\,\mathrm{e}^{-2\,x} \sin{(3\,x)} - \frac{2}{13}\,\mathrm{e}^{-2\,x} \cos{ \left( 3\,x \right) } + C \end{align*}$
 
Mathematics news on Phys.org
Another approach would be to consider:

$$y=\int e^{ax}\cos(bx)\,dx$$

Thus:

$$\d{y}{x}=e^{ax}\cos(bx)$$

The homogeneous solution is:

$$y_h(x)=c_1$$

And the particular solution will take the form:

$$y_p(x)=e^{ax}\left(A\cos(bx)+B\sin(bx)\right)$$

Hence:

$$y_p'(x)=e^{ax}\left((Aa+Bb)\cos(bx)+(Ba-Ab)\sin(bx)\right)$$

Substituting into the ODE, we obtain:

$$e^{ax}\left((Aa+Bb)\cos(bx)+(Ba-Ab)\sin(bx)\right)=e^{ax}\cos(bx)$$

$$(Aa+Bb)\cos(bx)+(Ba-Ab)\sin(bx)=\cos(bx)+0\sin(bx)$$

Equating coefficients, we obtain the system:

$$Aa+Bb=1$$

$$Ba-Ab=0$$

Solving this system, we obtain:

$$(A,B)=\left(\frac{a}{a^2+b^2},\frac{b}{a^2+b^2}\right)$$

And so our particular solution is:

$$y_p(x)=\frac{e^{ax}}{a^2+b^2}\left(a\cos(bx)+b\sin(bx)\right)$$

And then by the principle of superposition, there results:

$$y(x)=y_h(x)+y_p(x)=c_1+\frac{e^{ax}}{a^2+b^2}\left(a\cos(bx)+b\sin(bx)\right)$$

And so we conclude:

$$\int e^{ax}\cos(bx)\,dx=\frac{e^{ax}}{a^2+b^2}\left(a\cos(bx)+b\sin(bx)\right)+C$$

In the given problem, we have:

$$a=-2,\,b=3$$

Plugging those in, we have:

$$\int e^{-2x}\cos(3x)\,dx=\frac{e^{-2x}}{13}\left(3\sin(bx)-2\cos(bx)\right)+C$$
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
I'm interested to know whether the equation $$1 = 2 - \frac{1}{2 - \frac{1}{2 - \cdots}}$$ is true or not. It can be shown easily that if the continued fraction converges, it cannot converge to anything else than 1. It seems that if the continued fraction converges, the convergence is very slow. The apparent slowness of the convergence makes it difficult to estimate the presence of true convergence numerically. At the moment I don't know whether this converges or not.
Back
Top